What is the difference between Photo-excitation and photo-ionization?
I've never used the words before but it seems somewhat clear just looking at each word.
Both words use photo which in most cases if fairly synonymous with light.
Excitation is probably a process that deals with the transfer of energy to something
Ionization is, well, what I'm sure you think it means
Nothing wiki wouldnt tell you, but anyway
Photo excitation is exciting an atom with light i.e. a photon is absorbed causing the atom to attain a higher energy state. An atom is by default, in its lowest possible E state. A photon with appropriate energy (using a laser, most commonly) hits an electron causing it to gain the energy of this photon ( E conservation) and become excited.
Photo ionization would be knocking an electron out of its orbit using a photon with a very high energy (Xrays) making an ion.
Well its not what I am asking, maybe I should have completed my question before. In my book it says that photo-excitation is absorption of a photon by an electron making it jump to the higher state. However it says that the photon must correspond to definite energy level. Say if the electron needs 1.1 eV to jump to 1st higher state and 2.5 eV to another, then a photon of 1.3 eV is useless and won't be absorbed.
On the other hand for for photo ionization process, a photon of energy at least equal to ionization energy of the electron (say k eV) is required. Any other higher energy of the photon equals to the kinetic energy gained by the electron.
My doubt is that why doesn't the same happen in the first case? That the electron takes 1.1 eV to shift its energy state and uses the remaining 0.2 eV to increase the kinetic energy? Does something prevent that? (like stable orbits or something similar?)
Because a particular state of an atom IS comprised of a potential + kinetic energy. So the energy that it absorbs is already "pre-measured". The 1.1. eV is a combination of both KE and PE for that state.
Separate names with a comma.